Cambridge IGCSE Mathematics - Additional 0606 — 2019 May/June Paper 2 · Variant 2

This document consists of 16 printed pages. DC (LK) 165275/2 © UCLES 2019 [Turn over * 3 0 3 4 6 8 0 6 6 3 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2019 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator READ THESE INSTRUCTIONS FIRST Write your centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education

2 0606/22/M/J/19 © UCLES 2019 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 2 4 2 Binomial Theorem (a + b)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A ∆ = 1 2 bc sin A

3 0606/22/M/J/19 © UCLES 2019 [Turn over 1 Given that ln sin y x x 2 = , find an expression for d d x y . [4] 2 Find the values of k for which the equation ( ) k x kx k 1 0 2 - + - = has real and distinct roots. [4]

4 0606/22/M/J/19 © UCLES 2019 3 (i) Given that x 2 - is a factor of ax x x 12 5 6 3 2 - + + , use the factor theorem to show that a 4 = . [2] (ii) Showing all your working, factorise x x x 4 12 5 6 3 2 - + + and hence solve x x x 4 12 5 6 0 3 2 - + + = . [4]

5 0606/22/M/J/19 © UCLES 2019 [Turn over 4 A circle has diameter x which is increasing at a constant rate of . cms 0 01 1 - . Find the exact rate of change of the area of the circle when cm x 6 = . [5]

6 0606/22/M/J/19 © UCLES 2019 5 (i) Express x x 5 15 1 2 - + in the form p x q r 2 + + ` j , where p, q and r are constants. [3] (ii) Hence state the least value of . x x 3 0 2 2 - + and the value of x at which this occurs. [2]

7 0606/22/M/J/19 © UCLES 2019 [Turn over 6 (a) State the order of the matrix 0 5 1 8 4 1 8 6 e o. [1] (b) A 2 1 4 3 = - - e o (i) Find A 1 - . [2] (ii) Hence, given that ABA = I, find the matrix B. [3]

8 0606/22/M/J/19 © UCLES 2019 7 (a) Solve ( ) lg x 3 0 2 - = . [2] (b) (i) Show that, for a 0 2 , ln ln ln a a a 1 ( ) sin x 2 5 + + b l may be written as ( ) sin x k 2 5 + + , where k is an integer. [3] (ii) Hence find ln ln ln d a a a x 1 ( ) sin x 2 5 + + c e d dd b l . [3]

9 0606/22/M/J/19 © UCLES 2019 [Turn over 8 (a) In the binomial expansion of a x 2 6 - b l , the coefficient of x3 is 120 times the coefficient of x5. Find the possible values of the constant a. [4] (b) (i) Expand ( )x 1 2 20 + in ascending powers of x, as far as the term in x3. Simplify each term. [2] (ii) Use your expansion to show that the value of 0.9820 is 0.67 to 2 decimal places. [2]

10 0606/22/M/J/19 © UCLES 2019 9 (a) Solve sin cos x x 6 13 1 2 - = for ° ° x 0 360 G G . [4]

11 0606/22/M/J/19 © UCLES 2019 [Turn over (b) (i) Show that, for y 2 2 1 1 r r - , tan tan y y 1 4 2 + can be written in the form sin a y, where a is an integer. [3] (ii) Hence solve tan tan y y 1 4 3 0 2 + + = for y 2 2 1 1 r r - radians. [1]

12 0606/22/M/J/19 © UCLES 2019 10 (a) Find the unit vector in the direction of i j 5 15 - . [2] (b) The position vectors of points A and B relative to an origin O are 3 5 - e o and 12 7 e o respectively. The point C lies on AB such that AC : CB is 2 : 1. (i) Find the position vector of C relative to O. [3]

13 0606/22/M/J/19 © UCLES 2019 [Turn over The point D lies on OB such that OD : OB is 1 : m and . DC 6 1 25 = e o. (ii) Find the value of m. [3]

14 0606/22/M/J/19 © UCLES 2019 11 The velocity, v m s−1, of a particle travelling in a straight line, t seconds after passing through a fixed point O, is given by v t 1 4 3 = + ` j . (i) Explain why the direction of motion of the particle never changes. [1] (ii) Showing all your working, find the acceleration of the particle when t 5 = . [3] (iii) Find an expression for the displacement of the particle from O after t seconds. [3] (iv) Find the distance travelled by the particle in the fourth second. [2]

15 0606/22/M/J/19 © UCLES 2019 [Turn over 12 (a) The functions f and g are defined by ( ) f x x 5 2 = - for x 1 2 , ( ) g x x 4 9 2 = - for x 0 2 . (i) State the range of g. [1] (ii) Find the domain of gf. [1] (iii) Showing all your working, find the exact solutions of ( ) gf x 4 = . [3] Question 12(b) is printed on the next page.

16 0606/22/M/J/19 © UCLES 2019 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. (b) The function h is defined by ( ) h x x 1 2 = - for x 1 G- . (i) State the geometrical relationship between the graphs of ( ) h y x = and ( ) h y x 1 = - . [1] (ii) Find an expression for ( ) h x 1 - . [3]

This document consists of 9 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2019 MARK SCHEME Maximum Mark: 80 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 9 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 9 MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 9 Question Answer Marks Partial Marks 1 Either: Quotient rule: ( ) d sin cos d = x x x soi B1 ( ) 2 d 2 ln d = x x x oe soi B1 ( )( ) ( ) ( ) 2 2 2 2 ln cos sin ln x their x x their x x   −     oe M1 ( ) ( ) ( ) 2 2 2 2 ln cos sin ln x x x x x   −     oe isw A1 Or: Product rule on 2 1 (sin )(ln )− = y x x ( ) d sin cos d = x x x soi B1 ( ) 2 1 2 2 d 2 (ln ) (ln ) d − − = − × x x x x B1 2 2 2 1 2 (sin ) (ln ) ( cos )(ln ) oe x their x x their x x − −   × − ×     + M1 2 2 2 1 2 (sin ) (ln ) (cos )(ln ) oe isw x x x x x − −   × − ×     + A1 2 2 4( 1)( ) − − − k k k oe B1 (5 4) − k k M1 Correct critical values 0, 0.8 oe A1 0, 0.8 < > k k oe A1 FT their critical values provided their 2 0 + + > ak bk c has positive a and there are 2 values; mark final answer If B1 M0 allow SC1 for a final answer of k > 0.8 oe

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 9 Question Answer Marks Partial Marks 3(i) Uses x = 2 as a root: 3 2 (2 ) 12(2 ) 5(2) 6 0 − + + = a M1 or 2 a −12 5 6 ↓ 2a −24+4a −38+8a a −12+2a −19+4a −32+8a = 0 Solves 8 48 10 6 0 − + + = a to find a = 4 A1 or solves −32 + 8a = 0 to find a = 4 If M0 then SC1 for 3 2 4(2 ) 12(2 ) 5(2) 6 0 − + + = or showing that the synthetic division with a = 4 results in a remainder of 0 3(ii) 2 ( 2)(4 4 3) − − − x x x B2 B1 for any two terms correct in quadratic factor Product of three correct linear factors: ( 2)(2 3)(2 1) − − + x x x B1 2, 1.5, 0.5 = = = − x x x oe B1 dep on all previous marks having been earned If B2 B0 then award SC1 for correct factorisation of correct quadratic factor leading to 3 correct roots 4 Either: 2 1 π 4 = A x oe, soi B1 d 2 π d 4 = A x x oe, soi B1 d 0.01 d = x t soi B1 2 π(6) 0.01 4 × M1 FT their d d A x when x = 6 0.03π or exact equivalent A1 mark final answer Or: 2 π = A r and 2 = x r soi B1 d 2π d = A r r oe soi B1 d 0.005 d = r t oe soi B1 2π(3) 0.005 × M1 0.03π or exact equivalent A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 9 Question Answer Marks Partial Marks 5(i) 2 5( 1.5) 10.25 − − x isw B3 B1 for each of p, q, r correct in correctly formatted expression; allow correct equivalent values If B0 then SC2 for ( ) 5 1.5 10.25 − − x or SC1 for correct values but other incorrect format 5(ii) 10.25 5 − their is least value when 1.5 = x their B2 STRICT FT their part (i); B1 STRICT FT for each 6(a) 2 × 4 or 2 by 4 B1 6(b)(i) (A−1 =) 3 4 1 1 2 2       isw B2 B1 for 1 2   ×     their or for 3 4 1 2   ×    k , where k is not 0 or 1 6(b)(ii) B = 13 20 1 5 8 4       oe isw B3 FT their A−1 provided B1 earned in (b)(i) B1 for the strategy of using A−1: B = A−1 A−1 soi or AB = A−1 soi or BA = A−1 soi or B = (A−1)2 and B1 for two or three elements of B correct or correct FT 7(a) lg (x2 – 3) = lg 1 soi M1 −2 and 2 A1 Implies M1 7(b)(i) Two separate terms in numerator: (sin(2 5))ln + x a or sin(2 5) log (sin(2 5))log + = + x a a a x a B1 Combines terms in numerator: or sin(2 5) ln +       x a a ln − a or 1 log log −= − a a a a B1 or (sin(2 5) 1)ln + − x a or ( ) sin(2 5) sin(2 5) 1 ln log ln + + −      = x x a a a a a sin(2 5) 1 + − x B1 dep all previous marks awarded; 7(b)(ii) 1 cos(2 5) 2 − + x + ( their −1) x (+ c) B3 FT their numerical k B2 for 1 cos(2 5) 2 − + x seen or B1 for cos(2 5), 0 + < a x a or for 1 cos(2 5) 2 + x or for 1 cos2 5 2 − + x seen

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 9 Question Answer Marks Partial Marks 8(a) 3 3 20 8   −   a x and 5 6 32   −   a x oe soi or 3 3 20 8     a x and 5 6 32     a x oe soi B2 B1 for either 3 3 20 8   −   a x or 5 6 32   −   a x oe or for 6 3 3 3 8   −   C a x and 6 5 5 32   −   C a x oe, or for 6 3 3 3 8     C a x and 6 5 5 32     C a x oe or for 3 3 20 8     a x and 5     ka x oe where k > 0 or for 3 3     ka x and 5 6 32     a x oe where k > 0 3 20 6 120 8 32 = × their a their a oe soi M1 [ ] 3 = ± a A1 8(b)(i) 2 3 1 40 760 9120 + + + x x x B2 B1 for three out of the four terms correct If B0 then SC1 for 2 3 1, 40 , 760 , 9120 x x x seen but not summed 8(b)(ii) 2 3 1 40( 0.01) 760( 0.01) 9120( 0.01) + − + − + − or 1 − 0.4 + 0.076 − 0.00912 oe leading to 0.66688 cao B2 or M1 for use of 0.01 = − x oe in their expansion seen or implied by e.g. 0.66688 without working or 1 − 0.4 + 0.076 − 0.00912 9(a) 2 6(1 cos ) 13cos 1 − − = x x oe B1 Solves or factorises their 3-term quadratic M1 70.5 and 289.5 A2 with no extras in range A1 for either, ignoring extras in range 9(b)(i) Numerator: Substitution of sin tan cos = y y y M1 Denominator: Substitution of 2 2 1 tan sec + = y y or substitution of 2 2 2 sin 1 tan 1 cos + = + y y y and correct rearrangement to 2 1 cos y oe M1 Correct completion to 4sin y cao A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 9 Question Answer Marks Partial Marks 9(b)(ii) −0.848[06…] rot to 3 or more figures B1 with no extras in range 10(a) 2 2 5 ( 15) + − seen M1 ( ) 1 5 15 5 10 − i j oe, isw A1 10(b)(i) 9 12       oe, soi B1 3 9 2 5 12 3 their       +       −       oe, soi or 12 9 1 7 12 3 their       −             oe, soi M1 9 3    A1 If B1 M0, award SC1 for a final answer of 9 3 −     −   oe 10(b)(ii) Forms a valid vector relationship using JJJG DC and e.g. JJJG OD or JJJG DB e.g. 9 6 3 1.25 their OD    = +       JJJG oe or 6 9 1 1.25 12 3 DB their     = + ×         JJJG oe M1 or = − JJJG JJJG JJJG DC OC OD 6 9 12 1 1.25 3 7 their λ      = −           soi or = − JJJG JJJG JJJG DC DB CB 6 12 9 1 1 1.25 7 12 3 their λ λ       − = − ×             soi Finds a correct proportion e.g. 4 = JJJG JJJG OB OD oe soi or 3 4 = JJJG JJJG OB DB oe soi A1 or solves a correct equation in λ e.g. 1 6 9 12 λ = − × λ = 4 A1 from a fully correct method 11(i) 0 ≠ v or 0 > v oe B1 11(ii) Differentiates : 4 4 3( 1)− × − + t oe, isw B2 B1 for 4 ( 1)− + k t with k ≠ −12 1 108 − oe or −0.00926 B1 11(iii) Integrates: 2 [ ] 2( 1) 2 − = − + + s t oe, isw B3 B2 for [ ] 2 2( 1)− − + + t c or B1 for [ ] 2 ( 1)− + + k t c with k ≠ −2

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 9 Question Answer Marks Partial Marks 11(iv) Finds their s from (iii) when t = 4 or when t = 3 or finds ( ) ( ) ( ) 4 2 3 2 2 2( 1) [ 2] 2(4 1) 2(3 1) their t their − − −   − + +   = − + −− + M1 9 200 or 0.045 A1 12(a)(i) g > −9 B1 12(a)(ii) x > 1 B1 12(a)(iii) [ ] ( ) 2 gf ( ) 4 5 2 9 x x = − − B1 2 100 80 38 0 x x − − = or ( ) 2 45 9 5 2 4 x + − = M1 [ ] 2 ( 80) ( 80) 4(100)( 38) 2(100) x −− ± − − − = leading to 4 3 6 10 + oe only or 1 54 2 5 4   +       or better only A1 12(b)(i) (They are) reflections (of each other) in (the line) y = x oe B1 12(b)(ii) 2 2 1 = + x y or 2 2 1 = + y x M1 [ ] 2 1 = ± + x y or [ ] 2 1 = ± + y x A1 2 1 − + x nfww A1